Solve ∫ _{-T_{2}}^T_{2}((-y^2+4)-y^2) wrt y

$$\int _ { - T _ { 2 } } ^ { T _ { 2 } } ( ( - y ^ { 2 } + 4 ) - y ^ { 2 } ) d y$$

Evaluate

$$-\frac{4T_{2}^{3}}{3}+8T_{2}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -y^{2}+4-y^{2}\mathrm{d}y$$

Integrate the sum term by term.

$$\int -y^{2}\mathrm{d}y+\int 4\mathrm{d}y+\int -y^{2}\mathrm{d}y$$

Factor out the constant in each of the terms.

$$-\int y^{2}\mathrm{d}y+\int 4\mathrm{d}y-\int y^{2}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{2}\mathrm{d}y with \frac{y^{3}}{3}. Multiply -1 times \frac{y^{3}}{3}.

$$-\frac{y^{3}}{3}+\int 4\mathrm{d}y-\int y^{2}\mathrm{d}y$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$-\frac{y^{3}}{3}+4y-\int y^{2}\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{2}\mathrm{d}y with \frac{y^{3}}{3}. Multiply -1 times \frac{y^{3}}{3}.

$$-\frac{y^{3}}{3}+4y-\frac{y^{3}}{3}$$

Simplify.

$$-\frac{2y^{3}}{3}+4y$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$-\frac{2}{3}T_{2}^{3}+4T_{2}-\left(-\frac{2}{3}\left(-T_{2}\right)^{3}+4\left(-1\right)T_{2}\right)$$

Simplify.

$$-\frac{4T_{2}^{3}}{3}+8T_{2}$$

Differentiate w.r.t. T_2

$$8-4T_{2}^{2}$$